Lemma 89.3.1. The functor $F$ (89.3.0.1) is an equivalence.
89.3 Strategy
Let $S$ be a scheme. Let $X$ be a decent algebraic space over $S$. Let $x_1, \ldots , x_ n \in |X|$ be pairwise distinct closed points. For each $i$ we pick an elementary étale neighbourhood $(U_ i, u_ i) \to (X, x_ i)$ as in Decent Spaces, Lemma 68.11.4. This means that $U_ i$ is an affine scheme, $U_ i \to X$ is étale, $u_ i$ is the unique point of $U_ i$ lying over $x_ i$, and $\mathop{\mathrm{Spec}}(\kappa (u_ i)) \to X$ is a monomorphism representing $x_ i$. After shrinking $U_ i$ we may and do assume that for $j \not= i$ there does not exist a point of $U_ i$ mapping to $x_ j$. Observe that $u_ i \in U_ i$ is a closed point.
Denote $\mathcal{C}_{X, \{ x_1, \ldots , x_ n\} }$ the category of morphisms of algebraic spaces $f : Y \to X$ which induce an isomorphism $f^{-1}(X \setminus \{ x_1, \ldots , x_ n\} ) \to X \setminus \{ x_1, \ldots , x_ n\} $. For each $i$ denote $\mathcal{C}_{U_ i, u_ i}$ the category of morphisms of algebraic spaces $g_ i : Y_ i \to U_ i$ which induce an isomorphism $g_ i^{-1}(U_ i \setminus \{ u_ i\} ) \to U_ i \setminus \{ u_ i\} $. Base change defines an functor
To reduce at least some of the problems in this chapter to the case of schemes we have the following lemma.
Proof. For $n = 1$ this is Limits of Spaces, Lemma 70.19.1. For $n > 1$ the lemma can be proved in exactly the same way or it can be deduced from it. For example, suppose that $g_ i : Y_ i \to U_ i$ are objects of $\mathcal{C}_{U_ i, u_ i}$. Then by the case $n = 1$ we can find $f'_ i : Y'_ i \to X$ which are isomorphisms over $X \setminus \{ x_ i\} $ and whose base change to $U_ i$ is $f_ i$. Then we can set
This is an object of $\mathcal{C}_{X, \{ x_1, \ldots , x_ n\} }$ whose base change by $U_ i \to X$ recovers $g_ i$. Thus the functor is essentially surjective. We omit the proof of fully faithfulness. $\square$
Lemma 89.3.2. Let $X, x_ i, U_ i \to X, u_ i$ be as in (89.3.0.1). If $f : Y \to X$ corresponds to $g_ i : Y_ i \to U_ i$ under $F$, then $f$ is quasi-compact, quasi-separated, separated, locally of finite presentation, of finite presentation, locally of finite type, of finite type, proper, integral, finite, if and only if $g_ i$ is so for $i = 1, \ldots , n$.
Proof. Follows from Limits of Spaces, Lemma 70.19.2. $\square$
Lemma 89.3.3. Let $X, x_ i, U_ i \to X, u_ i$ be as in (89.3.0.1). If $f : Y \to X$ corresponds to $g_ i : Y_ i \to U_ i$ under $F$, then $Y_{x_ i} \cong (Y_ i)_{u_ i}$ as algebraic spaces.
Proof. This is clear because $u_ i \to x_ i$ is an isomorphism. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$
). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)